SOLUTION: volume of a square based box is 256cm^3. 1.)Find h in terms of l. if the box has an opened top, find: 2.)the surface area, A in terms of l only 3.) dimensions of the box if the

Algebra ->  Test -> SOLUTION: volume of a square based box is 256cm^3. 1.)Find h in terms of l. if the box has an opened top, find: 2.)the surface area, A in terms of l only 3.) dimensions of the box if the      Log On


   



Question 352094: volume of a square based box is 256cm^3.
1.)Find h in terms of l.
if the box has an opened top, find:
2.)the surface area, A in terms of l only
3.) dimensions of the box if the surface area is to be a minimum
4.) the minimum area.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
volume of a square based box is 256cm^3.
1.)Find H in terms of L.


V+=+LWH

Since the base (horizontal cross section) is square, L+=+W

V+=+L%28L%29H%0D%0A%7B%7B%7BV=L%5E2H

The volume is given as 256, so

256+=+L%5E2H

256%2FL%5E2=H

H=256%2FL%5E2


if the box has an opened top, find:
2.)the surface area, A in terms of L only

A rectangular solid has 6 sides, but without a top
it has only 5 sides, 4 of them have area LH and the
bottom is a square with area L² like this, but with 
the 4 rectangular sides (flaps) folded upward and taped 
to form a box:



A=4LH+%2B+L%5E2

And since H=256%2FL%5E2

A=4L%2Aexpr%28256%2FL%5E2%29+%2B+L%5E2

A=4cross%28L%29%2Aexpr%28256%2FL%5Ecross%282%29%29+%2B+L%5E2

A=4%2Aexpr%28256%2FL%29+%2B+L%5E2

A=1024%2FL+%2B+L%5E2


3.) dimensions of the box if the surface area is to be a minimum



We find the derivative of A with respect to L

A=1024%2FL+%2B+L%5E2

Rewrite it as

A=1024L%5E%28-1%29+%2B+L%5E2

dA%2FdL=-1024L%5E-2%2B2L

dA%2FdL=-1024%2FL%5E2%2B2L

We set that equal to 0 to find candidates for x values at which
A might be a minimum or a maximum:

-1024%2FL%5E2%2B2L=0

Multiply through by L%5E2 to clear of fractions:

-1024%2B2L%5E3=0

2L%5E3=1024

L%5E3=512

L=root%283%2C512%29

L=8

We need to determine if that is a maximum, minimum or neither.

Using 2nd derivative test:

dA%2FdL=-1024L%5E-2%2B2L

%28d%5E2A%29%2F%28dL%5E2%29=2048L%5E-3%2B2

%28d%5E2A%29%2F%28dL%5E2%29=2048%2FL%5E3%2B2

Substituting L=8

%28d%5E2A%29%2F%28dL%5E2%29=2048%2F8%5E3%2B2

%28d%5E2A%29%2F%28dL%5E2%29=4%2B2

%28d%5E2A%29%2F%28dL%5E2%29=6

That is positive so graph of A is concave upward
when L=8, so A has a minimum value when L=8

Since H=256%2FL%5E2

H=256%2F8%5E2=256%2F64=4

So the box has an 8cm² by 8cm² base with height 4cm.

4.) the minimum area.

Substitute in

A=4LH+%2B+L%5E2

A=4LH+%2B+L%5E2=4

A=4%288%29%284%29+%2B+%288%29%5E2=128%2B64=192

So the minimum outside surface area is 192cm²

Edwin